pset3sol - 22.02 Problem Set 3 Solution 1.Solution The...

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22.02 Problem Set 3 Solution 1.Solution The energy eigenvalue problem for the given system is: H ˆ ψ = E ψ h 2 d 2 H ˆ = − 2 m dx 2 + V ( x ) 0 | x | > a V ( x ) = V 0 | x | a For bounded particle, the solution is: ψ = Ae κ x x < − a 1 ψ 2 = B sin kx or B 'cos kx a x a ψ 3 = Ce κ x x > a k = 2 0 ) ( 2 h V E m + , κ = 2 2 h mE Since the solution has to be odd, ψ 2 = B sin kx Also we know, ψ 2 ( a ) = ψ 3 ( a ) , if ψ 3 ( x ) is decaying, there has to be at least a quarter of wave from 0 to a within the well. This means λ / 4 a . “Just being bounded” means λ / 4 = a and E = 0 . This gives us k = m a V mV a 2 2 2 0 2 0 8 2 4 2 2 h h π π λ π = = = 2.Solution Liboff 5.12 (a) Proof [ A ˆ + B ˆ , C ˆ ] = [ A ˆ , C ˆ ] + [ B ˆ , C ˆ ] LHS = [ A ˆ + B ˆ , C ˆ ] = ( A ˆ + B ˆ ) C ˆ C ˆ ( A ˆ + B ˆ ) = A ˆ C ˆ + B ˆ C ˆ C ˆ A ˆ C ˆ B ˆ RHS = [ A ˆ , C ˆ ] + [ B ˆ , C ˆ ] = A ˆ C ˆ C ˆ A ˆ + B ˆ C ˆ C ˆ B ˆ = LHS ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (b) Proof [ AB , C ] = A [ B , C ] + [ A , C ] B ˆ ˆ ˆ ˆ ˆ ˆ LHS = ABC CAB RHS = A ˆ ( B ˆ C ˆ C ˆ B ˆ ) + ( A ˆ C ˆ C ˆ A ˆ ) B ˆ = A ˆ B ˆ C ˆ A ˆ C ˆ B ˆ + A ˆ C ˆ B ˆ C ˆ A ˆ B ˆ = A ˆ B ˆ C ˆ C ˆ A ˆ B ˆ = LHS Liboff 5.19 ( ϕ 2 , ϕ 3 ) = {sin kx ,exp( ikx )} Since we know ϕ 3 = exp( ikx ) is a common eigenfunction of H ˆ and p ˆ , we’d like to keep it as one of the two common eigenfunctions composed of ( ϕ 2 , ϕ 3 ) . If we denote it as ϕ a , then ϕ a = a 2 ϕ 2 + a 3 ϕ 3 = ϕ 3 , which means a 2 = 0 and a 3 = 1 . (1) Also we know ϕ b = exp( ikx ) is a common eigenfunction of H ˆ and p ˆ and is linearly independent of ϕ a = exp( ikx ) , we hope we could construct ϕ b as a linear combination of ( ϕ 2 , ϕ 3 ) . ϕ b = b 2 ϕ 2 + b 3 ϕ 3 = b 2 sin kx + b 3 exp( ikx ) = b 2 sin kx + b 3 (cos kx + i sin kx )
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22.02 Problem Set 3 Solution ϕ b = b 3 cos kx + ( b 2 + ib 3 )sin kx = exp( ikx ) = cos kx i sin kx b 3 = 1 , b 2 = − 2 i (2) Sum (1) and (2), we get: ϕ a = 0 × ϕ 2 + 1 × ϕ 3 = exp( ikx ) ϕ b = − 2 i × ϕ 2 + 1 × ϕ 3 = exp( ikx ) ϕ 2 and ϕ 3 are a pair of linearly independent common eigenfunctions of H ˆ and p ˆ . Liboff 5.25 Use uncertainty principle: If [ A ˆ , B ˆ ] = C 0 , then A B 1 | < C > | 2 (a) x E [ x ˆ, E ˆ ] ψ = x ˆ E ˆ ψ E ˆ x ˆ ψ h 2 d 2 x ˆ = x , E ˆ = − 2 m dx 2 + V ( x ) h 2 d 2 ψ h 2 d 2 [ x ˆ, E ˆ ] ψ = x [ 2 m dx 2 + V ( x ) ψ ] [ 2 m dx 2 + V ( x )][ x ψ ] h 2 d 2 ψ h 2 d 2 = − x 2 m dx 2 + xV ( x ) ψ + 2 m dx 2 [ x ψ ] xV ( x ) ψ h 2 d 2 ψ h 2 d d ψ h 2 d 2 ψ h 2 d ψ h 2 d 2 ψ h 2 d ψ + + + x + = − x 2 m dx 2 2 m dx [ ψ + x dx ] = − x 2 m dx 2 2 m dx 2 m dx 2 2 m dx [ x ˆ, E ˆ ] = h 2 d m dx d p ˆ x = − i h dx [ x ˆ, E ˆ ] = i h p ˆ x m 1 h From uncertainty principle, we get x E 2 | < i p ˆ x > | m Thus x E 1 h | < p ˆ x > | 2 m (b) p x E d h 2 d 2 p ˆ x = − i h dx , E ˆ = − 2 m dx 2 + V ( x ) [ p ˆ x , E ˆ ] ψ = p ˆ x E ˆ ψ E ˆ p ˆ x ψ
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22.02 Problem Set 3 Solution [ p ˆ x , E ˆ ] ψ = i h d [ h 2 d 2 ψ + V ( x ) ψ ] [ h 2 d 2 2 + V ( x )][ i h d ψ ] dx 2 m dx 2 2 m dx dx h 3 = i d 3 ψ i h dV ( x ) ψ i h d ψ V ( x ) i h 2 d 3 ψ + i h V ( x ) d ψ 2 m dx 3 dx dx 2 m dx 3 dx = − i h dV ( x ) ψ dx p x E 1 | < − i h dV ( x ) > | = h | < dV ( x ) > | 2 dx 2 dx (c) x T h 2 d
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